What is expand (e^xcos(2x))(e^xsin(2x)+2e^xcos(2x)) ?

The solution to expand (e^xcos(2x))(e^xsin(2x)+2e^xcos(2x)) is e^{2x}sin(2x)cos(2x)+2e^{2x}cos^2(2x)

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expand (e^xcos(2x))(e^xsin(2x)+2e^xcos(2x))

expand

(e^{x} cos (2 x)) (e^{x} sin (2 x) + 2 e^{x} cos (2 x))

e^{2 x} sin (2 x) cos (2 x) + 2 e^{2 x} cos^{2} (2 x)

Solution steps

(e^{x} cos (2 x)) (e^{x} sin (2 x) + 2 e^{x} cos (2 x))

Apply the distributive law:

a (b + c) = ab + a c

a = (e^{x} cos (2 x)), b = e^{x} sin (2 x), c = 2 e^{x} cos (2 x)

= (e^{x} cos (2 x)) e^{x} sin (2 x) + (e^{x} cos (2 x)) \cdot 2 e^{x} cos (2 x)

= e^{x} sin (2 x) (e^{x} cos (2 x)) + 2 e^{x} cos (2 x) (e^{x} cos (2 x))

Simplify

e^{x} sin (2 x) (e^{x} cos (2 x)) + 2 e^{x} cos (2 x) (e^{x} cos (2 x)) : e^{2 x} sin (2 x) cos (2 x) + 2 e^{2 x} cos^{2} (2 x)

= e^{2 x} sin (2 x) cos (2 x) + 2 e^{2 x} cos^{2} (2 x)

e^{7 + i (- \frac{π}{2} + 2 kπ)}

(m + p = 5, 5) (f (p) - p = 2, 7) (f (p) + m = 28)

\frac{2 z}{( 2 z - 1 ) ( 4 z - 1 )}

(4 + 6 i)^{8}

\frac{6 + s ^{5} - 4 s ^{4}}{s ^{4} ( s - 2 ) ^{2}}

What is expand (e^xcos(2x))(e^xsin(2x)+2e^xcos(2x)) ?
The solution to expand (e^xcos(2x))(e^xsin(2x)+2e^xcos(2x)) is e^{2x}sin(2x)cos(2x)+2e^{2x}cos^2(2x)